2025 · Q41
If p sin^2α + q cos^2α = m and q sin^2β + p cos^2β = n, then what is (tanα/tanβ)^2 equal to?
Elementary MathematicsTrigonometry
- A(m-p)(n-q)/((m-q)(n-p))
- B(m-q)(n-q)/((m-p)(n-p))
- C(m-q)(n-p)/((m-p)(n-q))
- D(m-p)(n-p)/((m-q)(n-q))
Explanation
From p*sin^2(alpha)+q*cos^2(alpha)=m: sin^2(alpha)=(m-q)/(p-q), cos^2(alpha)=(p-m)/(p-q), so tan^2(alpha)=(m-q)/(p-m). From q*sin^2(beta)+p*cos^2(beta)=n: tan^2(beta)=(p-n)/(n-q). Hence (tan(alpha)/tan(beta))^2=(m-q)(n-q)/((m-p)(n-p)) - option B. (Option C gives a negative value for valid inputs, e.g. p=5,q=2, so it cannot be a squared ratio.) The stored key C was wrong; corrected to B.